Question

5. Consider the process where nA mol of gas A initially at 1 bar pressure mix with nB mol of gas B also at 1 bar to form 1 mol of a uniform mixture of A and B at a final total pressure of 1 bar, and all at constant temperature T. Assume that all gases behave ideally. a. Show that the entropy change, mixSm, for this process is given by mixSm

Answer 1

Answer:

ΔSmix,m = 5.7628 J/K.mol

Explanation:

mix: A + B

∴ nA = x mol A

∴ nB = y mol B

⇒ n mix = x + y = 1 mol

∴ P total = 1 bar

∴ T: constant

entropy of gases when mixing:

• ΔSmix = - nA*R*LnXA - nB*R*LnXB

∴ XA = x/1 = x

∴ XB = y/1 = y

⇒ ΔSmix = - x*R*Lnx - y*R*Lny

assuming: x = y = 0.5 mol

⇒ ΔSmix = - (0.5)(R)(- 0.693) - (0.5)(R)(- 0.693)

⇒ ΔSmix = (0.3465)(R) + (0.3465)(R)

⇒ ΔSmix = (0.6931)(R)

∴ R = 8.314 J/K.mol

⇒ ΔSmix,m = (0.6931)(8.314 J/K.mol)

⇒ ΔSmix,m = 5.7628 J/K.mol